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3 years ago

9/10 ÷ 2/10 ansewer

Mathematics

2 answers:

muminat3 years ago

5 0

Answer: 4.5 or 9/2

Step-by-step explanation:

9/10 ÷ 2/10

9/10 x 10/2

9 x 10 = 90

10 x 2 = 20

Simplified: 9/2 or 4.5

k0ka [10]3 years ago

3 0

Answer:

4.5 or 4 and a half

Step-by-step explanation:

9/10 divided by 2/10

put 2/10 into it's reciprocal and multiply it by 9/10

10/2 * 9/10

10 9

_ * _

2 10

^ (visualization)

multiple into 90/20, divide numerator and denominator by 10 to get 9/2

9/2 is 4.5

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I need help with all of thies

levacccp [35]

Answer:

The equation for finding the hypotenuse is:

c^2+c^2=h^2

For example, in the first exercise:

8^2+10^2=164^2

For clearing the equation, find out the root of 164=

12.8

Hypotenuse of the first triangle is 12.8!

Let me help you with the next, if you still dont get it:

8^2+13^2=233^2

Root of 233: 15.2

Hypotenuse of second triangle is 15.2!

3 0

2 years ago

E price of a large pizza increased from $21 to $26. what is the percent of increase in the cost of pizza? round to the nearest p

rodikova [14]

Percent increase=increase/original times 100

increase=26-21=5
original=21

so
5/21 times 100=0.23809523809523809523809523809524 times 100=23.809523809523809523809523809524
round
24%

so 24% increase

7 0

3 years ago

The joint probability density function of X and Y is given by fX,Y (x, y) = ( 6 7 x 2 + xy 2 if 0 < x < 1, 0 < y < 2

fredd [130]

I'm going to assume the joint density function is

$f_{X,Y}(x,y)=\begin{cases}\frac67(x^2+\frac{xy}2\right)&\text{for }0$

a. In order for $f_{X,Y}$ to be a proper probability density function, the integral over its support must be 1.

$\displaystyle\int_0^2\int_0^1\frac67\left(x^2+\frac{xy}2\right)\,\mathrm dx\,\mathrm dy=\frac67\int_0^2\left(\frac13+\frac y4\right)\,\mathrm dy=1$

b. You get the marginal density $f_X$ by integrating the joint density over all possible values of $Y$ :

$f_X(x)=\displaystyle\int_0^2f_{X,Y}(x,y)\,\mathrm dy=\boxed{\begin{cases}\frac67(2x^2+x)&\text{for }0$

c. We have

$P(X>Y)=\displaystyle\int_0^1\int_0^xf_{X,Y}(x,y)\,\mathrm dy\,\mathrm dx=\int_0^1\frac{15}{14}x^3\,\mathrm dx=\boxed{\frac{15}{56}}$

d. We have

$\displaystyle P\left(X$

and by definition of conditional probability,

$P\left(Y>\dfrac12\mid X\frac12\text{ and }X$

$\displaystyle=\dfrac{28}5\int_{1/2}^2\int_0^{1/2}f_{X,Y}(x,y)\,\mathrm dx\,\mathrm dy=\boxed{\frac{69}{80}}$

e. We can find the expectation of $X$ using the marginal distribution found earlier.

$E[X]=\displaystyle\int_0^1xf_X(x)\,\mathrm dx=\frac67\int_0^1(2x^2+x)\,\mathrm dx=\boxed{\frac57}$

f. This part is cut off, but if you're supposed to find the expectation of $Y$ , there are several ways to do so.

Compute the marginal density of $Y$ , then directly compute the expected value.

$f_Y(y)=\displaystyle\int_0^1f_{X,Y}(x,y)\,\mathrm dx=\begin{cases}\frac1{14}(4+3y)&\text{for }0$

$\implies E[Y]=\displaystyle\int_0^2yf_Y(y)\,\mathrm dy=\frac87$

Compute the conditional density of $Y$ given $X=x$ , then use the law of total expectation.

$f_{Y\mid X}(y\mid x)=\dfrac{f_{X,Y}(x,y)}{f_X(x)}=\begin{cases}\frac{2x+y}{4x+2}&\text{for }0$

The law of total expectation says

$E[Y]=E[E[Y\mid X]]$

We have

$E[Y\mid X=x]=\displaystyle\int_0^2yf_{Y\mid X}(y\mid x)\,\mathrm dy=\frac{6x+4}{6x+3}=1+\frac1{6x+3}$

$\implies E[Y\mid X]=1+\dfrac1{6X+3}$

This random variable is undefined only when $X=-\frac12$ which is outside the support of $f_X$ , so we have

$E[Y]=E\left[1+\dfrac1{6X+3}\right]=\displaystyle\int_0^1\left(1+\frac1{6x+3}\right)f_X(x)\,\mathrm dx=\frac87$

5 0

3 years ago

Find the measure of the missing angle.

joja [24]

180-156= 24

24°!!
hope this helped u<3

4 0

2 years ago

Read 2 more answers

Answer this for me please

dybincka [34]

Answer:

-3

Step-by-step explanation:

You cant count the spaces up (9) and over (-3) between the two points and do up/over so it would be 9/-3 and simplified that is -3/1 or 3

6 0

2 years ago

9/10 ÷ 2/10 ansewer​

9/10 ÷ 2/10 ansewer